IIT JEE , AIEEE Forum - Mathematics , Algebra

tarinbansal

SO guys, I have started a new thread for solutions of Maths QBANK. All the authors OR anyone who has solved any question is invited to post their solutions here.

But please keep in mind to write the question before posting the solution as the questions in the QBANK are not numbered.

Don't think that U have to write the whole question here again to post its solution. Just copy paste it here from QBANK.

Hope it helps all the GOIITians in the New Year.

SO keep answering.

tarinbansal

So lets start.

The solution to my first question is coming up-

Q.1- Find the minimum value of (u-v)² + [ (2- u²)^1/2 - 9/v]² for 0<u<

2, v>0?

All right, it is a very good question and the only one(as far as I know) question of its type. First of all it must strike to U that it is a question of COORDIANTE GEOMETRY NOT Algebra.

Now to solve it- Did U notice that the given equation is actually the distance between 2 points in the coordinate plane.

Now when U have noticed that, U must also notice that their abscissae lie on the circle x² + y² = 2 and their ordinates lie on the rectangular hyperbola xy=9.

Hence, actually in the question we are asked the square of shortest distance between the above written circle and hyperbola which will be along the common normal to both of them.

Now be symmetry, one of the common normal will be the line y=x. Its intersection points on circle and hyperbola are (1,1) and (3,3) respectively.

Hence we have to find (3-1)² + (3-1)²

That is = 8.

Hence Ans=8

Hope U find this problem really gud and the solution satisfactory.

P.S.-Will keep posting solutions to other questions at regular intervals. If U want any solution immediately OR earlier than other questions, then nudge me about that question and I will post its solution.

budku007

If P(3,1) ,Q(6,5) and R(x,y) are 3 points such that anglePRQ=90 and areaPRQ=7, then the no of such points R=

solution

angle PRQ=90 degree

using m₁m₂=-1

y-1/x-3*y-5/x-6=-1

on solving we get locus of this point as x²+y²-9x-6y+23=0

angle PRQ is 90

so Triangle is right angled

so area is 1/2*base*height

find length of base and height

and using formula and equating with 7 we get another equation of this point as x²+y²-9x-6y-125/2=0

Now the point will be the intersection of these two loci

we see that both circles are concentric and distinct,so no common point

Hence no point.

Rate me if satisfied

tarinbansal

The equation x² + y² + 4x + 6y + 13=0 represents a/an __________?

The above equation can be re-written as (x+2)² + (y+3)²=0

Now the sum of 2 squared quantities can be zero only when they both are equal to zero.

Hence x=-2,y=-3

Hence the equation represents the point (-2,-3).

tarinbansal

In a ABC, 1/(a+b) + 1/(b+c) = 3/(a+b+c). Then B = ?

In the given equation, after taking LCM etc and solving we get-

a² - b² + c² -ac=0

Hence a² + c² -ac= b² --------------------- (1)

Now applying cosine formula for triangles-

a² + c² -2ac cosB= b² ----------------------(2)

From equations (1) and (2), we get cosB=1/2

Hence B = /3

rhd92781

Solution to Q. 3)

a,b,c are the cube roots of p, (p<0). Then for any permissible value of x,y,z which is given by

|xa+yb+zc| + (a₁²- 2b₁²) w + w²([x]+[y]+[z]) = 0

|xb+yc+za|

(where w is cube root of unity and a1,b1 are real positive number and prime number), find the value of [x+a₁] + [y+b₁] + [z], where [.] denotes the greatest integer function.

Sol) let x³=p(cos2n

+ isin2n

)

or x=p^1/3(cos2n

/3 + isin2n

/3)

Putting n=1, 2, 3 here, we get a=p^1/3,b=p^1/3w, c=p^1/3w²

(where w is the cube root of unity)

Now, |(xa+yb+zc)/(xb+yc+za)| = |p^1/3(x+yw+zw²)/p^1/3(xw+yw²+z)|

=|1/w||(xw+yw²+z)/(xw+yw²+z)|

=1

So we have,

1+(a1²-2b1²)w + w²([x]+[y]+[z]) = 0

This is possible only if a1²-2b1²=1 & [x]+[y]+[z]=1

Here a1=3, b1=2 ....(how? see this: http://www.goiit.com/posts/list/differenciation-tricky-question-33029.htm#166728)

now, [x+a1] + [y+b1] + [z] = [x]+[y]+[z] + a1+b1...(since a1, b1 are integers)

= 1 + 3 + 2 = 6

chintan_zalani

hey tarin this equation (x+2)² + (y+3)²=0 can represent "a point circle also with centre (-2,-3)" don't u think so???

tarinbansal

Point circle and point are one and the same thing. No diff between them. The diff lies in the point of view of diff people.

See I did not made that question. I also saw that in a book and I thought that its a gud question so posted here. The answer given in the book was a point.

NEways, point and a point circle, hardly makes any difference.

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